Question

A certain p-n junction, having a depletion region of width 20mm was found to have a breakdown voltage of 100V. If the width of the depletion region is reduced to 1 mm during its production, then it can be used as a Zener diode for voltage regulation of

• A. 15V
• B. 5V
• C. 7.5V
• D. 2000V

Hint:

Remember that for Zener diodes, the breakdown voltage is inversely proportional to the width of the depletion region. Reducing the width increases the breakdown voltage.

Answer 1

The Correct Option is B

Step 1: For a Zener diode, the breakdown voltage \( V \) is directly proportional to the depletion-region width \( W \), since a narrower depletion region reaches the critical breakdown field at a lower applied voltage: \( V \propto W \).
Step 2: Using the given data, \( V_1 = 100 \) V corresponds to \( W_1 = 20 \) mm, so the new voltage scales in the same ratio as the new width: \[ V_2 = V_1 \times \dfrac{W_2}{W_1} = 100 \times \dfrac{1}{20} \]
Step 3: Simplifying this gives the new Zener (regulation) voltage. \[ \boxed{V_2 = 5\text{V}} \]

Answer 2

A third way to approach this is with a pure scaling-factor argument: we don't need to derive the underlying physical law at all, only to notice by how much the depletion width has shrunk and apply the same shrink factor to the voltage, since the two quantities scale together in this problem.

The width shrinks from 20 units to 1 unit, a reduction by a factor of \[ \dfrac{W_1}{W_2} = \dfrac{20}{1} = 20 \] Since voltage scales down by this same factor of 20 (matching the direct proportionality between breakdown voltage and depletion width), the new voltage is \[ V_2 = \dfrac{V_1}{20} = \dfrac{100}{20} = 5\text{V} \]

1. 15V: Does not correspond to dividing 100V by the scaling factor of 20.
2. 5V: Exactly matches \( 100/20 \), consistent with the scaling-factor argument.
3. 7.5V: Not obtainable from this factor-of-20 scaling.
4. 2000V: This would arise only from multiplying (not dividing) by the scaling factor, which would incorrectly suggest that narrowing the depletion region raises the breakdown voltage, the opposite of the real physical trend.

Therefore, the correct answer is 5V.